Let $F$ be a free group, and $w$ an element of $F$. In any group $G$, a $w$-word is the image of $w$ or $w^{-1}$ under a homomorphism from $F$ to $G$. The subgroup of $G$ generated by $w$-words is denoted $G(w)$.

For any $g \in G(w)$, the $w$-length of $g$, denoted $l(g|w)$, is the minimum number of $w$-words in $G$ whose product is $g$, and the stable $w$-length of $g$, denoted $sl(g|w)$, is the limit $sl(g|w) = lim_{n \to \infty} l(g^n|w)/n$.

If $w$ is not in the commutator subgroup of $F$, the stable $w$-length of every element in any group is trivial. Otherwise, one has a universal inequality
$$1/2 \le sl_F(w|w) \le 1$$
(where the subscript $F$ indicates that stable $w$-length is being calculated in the free group $F$ containing $w$ itself.)

The lower bound of $1/2$ is realized e.g. by the word $w=xyx^{-1}y^{-1}$ (i.e. a standard commutator) in $F_2$ but I don't know how to compute (or even approximate!) $sl(w|w)$ in (essentially) any other case.

What values are achieved by $sl(w|w)$? Are they all rational? Are they dense? Is $1$ ever achieved? Is $1/2$ ever achieved for a word other than $xyx^{-1}y^{-1}$?

(Added:) After reading FC's answer, it is probably worth pointing out that the lower bound $1/2 \le sl_F(w|w)$ comes from the inequality $scl_G(g) \le sl_G(g|w)(scl_F(w)+1/2)$ for any $g$ in any $G$, so one gets a better lower bound on $sl_F(w|w)$ if one knows $scl_F(w)>1/2$ (the estimate $scl_F(w)\ge 1/2$ is always true). Upper bounds can be established by exhibiting identities (like FC's identity below). Does a blind computer search yield any interesting examples?

An interesting question! Can you give some indication of the argument showing that the stable $w$-length of every element in any group is trivial if $w$ is not in the commutator subgroup of $F$?
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Andy PutmanNov 11 '09 at 3:30

1

Hi Andy - the argument is actually trivial: if $w$ is not in the commutator subgroup of $F$, then there is a homomorphism from $F$ to $Z$ sending $w$ to some nonzero $n$. This means that the $n$th power of any element (and therefore the $nm$th power for any $m$) in any group is a $w$-word, so the stable $w$ length of anything is zero.
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Danny CalegariNov 11 '09 at 3:39

2 Answers
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Here are some weak observations that don't quite answer any of your questions. Let $g$ be a positive integer, and consider the free group $F_{2g}$ generated by $a_k$ and $b_k$ for $k = 1$ to $g$. Consider the word:

$$w_g = [a_1,b_1][a_2,b_2][a_3,b_3] \ldots [a_g,b_g].$$

Suppose that $\lambda_g = sl(w_g,w_g)$. I claim that for any $x$ in the commutator of $F_2$ with $cl(x) = g$, the stable commutator length $scl(x)$ of $x$ is $\le g \cdot \lambda_g$. Suppose otherwise. First of all, note that for large $n$ we can write $w^n_g$ as the product of (roughly) $n \cdot \lambda_g \cdot g$ commutators. Since the commutator length of $x$ is $g$, there exists a map from $F_{2g}$ to $F_{2}$ such that the image of $w_g$ is $x$. On the other hand, we see that the image of $w^n_g$ is $x^n$, and thus the commutator length of $x^n$ is (asymptotically) at most by $n \cdot \lambda_g \cdot g$, and thus $scl(x) \le \lambda_g \cdot g$.

Example: $cl([x,y]^3) = 2$ and $scl([x,y]^3) = 3/2$, and thus $\lambda_3 \ge 3/4$. In general, the fact that $scl([x,y]) = 1/2$ implies that that $\lambda_g$ tends to one as $g$ increases.

I think one can promote this example to a word in $F_2$. Consider the characteristic homomorphism $\phi_n:F_2 \rightarrow \mathbf{Z} \oplus \mathbf{Z}
\rightarrow \mathbf{Z}/n\mathbf{Z} \oplus \mathbf{Z}/n\mathbf{Z}$. Suppose that $n$ is odd, and write $2g = n^2 + 1$. The kernel of $F_2$ is free of rank $2g$. Pick generators for $\ker(\phi_n)$ once and for all, and call them $a_k$ and $b_k$ for $k = 1$ to $g$. We may think of $a_k$ and $b_k$ as elements in $F_2$, but also as formal words. Since $\ker(\phi_n) = F_{2g}$ is characteristic, the formal words $a_k$ and $b_k$ always yield elements of $F_{2g}$ (alternatively, the images of $a_k$ and $b_k$ in $\mathbf{Z} \oplus \mathbf{Z}$
are divisible by $n$, and this will be so for any substitution of elements of $F_2$ for the generators). Let

$$w_g = [a_1,b_1][a_2,b_2] \ldots [a_g,b_g].$$

The argument proceeds as above. If $sl(w_g,w_g) = \mu_g$, then we can write $w^n_g$ (for large $n$) as the product of $n \cdot \mu_g \cdot g$ commutators, each of which is the commutator of a pair of elements of $F_{2g}$ (by the characteristic property of the words $a_k$ and $b_k$ described above). Hence, choosing an appropriate map from $F_{2g}$ to $F_2$, we may deduce that for any $x \in F_2$ with $cl(x) = g$ that $scl(x) \le g \cdot \mu_g$. Thus we have found words $w_g$ in $F_2$ such that $sl(w_g,w_g)$ tends to $1$ as $g$ goes to infinity. Of course, this says nothing about whether $sl(w_g,w_g)$ actually equals $1$ for any $g$.

Hi FC - very nice answer! Let $a_1,b_1,\cdots,a_g,b_g$ generate a free group $F_g$. Let $w_g = [a_1,b_1]\cdots[a_g,b_g] \in F_g$. Then $scl(w_g) = g-1/2$ exactly in $F_g$, so $sl(w_g|w_g)=1-1/2g$ exactly, so that proves rationality for these elements. I know where you found your formula for $w^3$, but I don't see a pattern . . .
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Danny CalegariNov 13 '09 at 4:45

In case any one is still thinking about this question, it turns out one can say a lot. For example, $sl(w|w)=1/2$ whenever $w$ is a word of the form $[a,b^n]$ (and several other examples), one has $2/3 \le sl(w|w) \le 4/5$ when $w=[a,b]^2$, and if $\gamma_n$ is the iterated commutator $[x_1,[x_2,\cdots[x_{n-1},x_n]\cdots]$ one has $sl(\gamma_n|\gamma_n)\le 1-2^{1-n}$. In fact, I would explicitly conjecture that $sl(w|w)<1$ (i.e. strict inequality) for every $w$.

Getting systematic lower bounds on $sl(w|w)$, other than $scl(w)/(scl(w)+1/2)$ seems difficult; one imagines that the (currently nonexistent) theory of nonabelian Bavard duality might do the trick.